1. ## Prime counting function

I'm trying to spark some interest in discussion in prime number theory and the Riemann Zeta Function/Hypothesis. The prime counting function, $\pi(x)$, outputs the number of primes less than or equal to x. Gauss suggested at 15 that a good approximation for this prime counting function was $Li(x)$, where $Li(x)=\int_{0}^{\infty}\frac{dx}{\log(x)}$ (I realize that log(1)=0). This is a good approximation and although it seems to always bound the prime counting function, they do eventually cross.

Anyway, here is an explicit formula for a prime counting function.

$J(x)=Li(x)-\sum_{\rho}\frac{x^{\rho}}{\rho}-\ln(2)+\int_{x}^{\infty}\frac{dt}{t(t^2-1}\ln(t)}$

Here $\rho$ represents the non-trivial zeros of the Riemann Zeta Function in the critical strip. Since there are infinitely many zeros in the critical strip this formula is not feasible to use for actual caculations, but my question is that if we could use this formula would it output integers? I mean would it say exactly how many primes are less than or equal to x?

I'd appreciate any other thoughts to add to my own.

2. Originally Posted by Jameson
I'm trying to spark some interest in discussion in prime number theory and the Riemann Zeta Function/Hypothesis.
Good luck with that.

Originally Posted by Jameson
where $Li(x)=\int_{0}^{\infty}\frac{1}{\log(x)}$
Perhaps you mean to say,
$\mbox{Li}(x)=\int^x_2\frac{du}{\log u}$

3. Speaking about the prime counting function there is an elemantary proof (using only basic analysis) in my book that
$\lim_{n\to \infty}\frac{\pi (n)}{n}=0$

4. I forgot the dx, but I meant what I wrote. I technically should use limit notation to avoid the asymptote in the function, but I assume that you can notice that. There are two common Li(x) functions it seems.

http://en.wikipedia.org/wiki/Logarithmic_integral

5. Originally Posted by ThePerfectHacker
Speaking about the prime counting function there is an elemantary proof (using only basic analysis) in my book that
$\lim_{n\to \infty}\frac{\pi (n)}{n}=0$
This is interesting. I'll look into it.